1(1994), No. 4, 343-351
ON THE CORRECTNESS OF LINEAR BOUNDARY VALUE PROBLEMS FOR SYSTEMS OF GENERALIZED
ORDINARY DIFFERENTIAL EQUATIONS
M. ASHORDIA
Abstract. The sufficient conditions are established for the
correct-ness of the linear boundary value problem
dx(t) =dA(t)·x(t) +df(t), l(x) =c0,
whereA: [a, b]→Rn×n
andf : [a, b]→Rn
are matrix- and vector-functions of bounded variation,c0∈Rn
, andlis a linear continuous operator from the space ofn-dimentional vector-functions of bounded variation intoRn
.
Let the matrix- and vector-functions,A: [a, b]→Rn×n andf : [a, b]→
Rn, respectively, be of bounded variation,c0 ∈Rn, and letl :BVn(a, b)→
Rn be a linear continuous operator such that the boundary value problem
dx(t) =dA(t)·x(t) +df(t), (1)
l(x) =c0 (2)
has the unique solutionx0.
Consider the sequences of matrix- and vector-functions of bounded vari-ation Ak : [a, b]→Rn×n (k= 1,2, . . .) andfk : [a, b]→Rn (k= 1,2, . . .), respectively, the sequence of constant vectorsck ∈Rn(k= 1,2, . . .) and the sequence of linear continuous operatorslk:BVn(a, b)→Rn (k= 1,2, . . .).
In this paper the sufficient conditions are given for the problem
dx(t) =dAk(t)·x(t) +dfk(t), (3)
lk(x) =ck (4)
to have a unique solutionxk for any sufficiently large kand lim
k→+∞xk(t) =x0(t) uniformly on [a, b]. (5)
1991Mathematics Subject Classification. 34B05. 343
An analogous question is studied in [2–4] for the boundary value problem for a system of ordinary differential equations.
The theory of generalized ordinary differential equations enables one to investigate ordinary differential and difference equations from the common standpoint. Moreover, the convergence conditions for difference schemes corresponding to boundary value problems for systems of ordinary differen-tial equations can be deduced from the correctness results for aprropriate boundary value problems for systems of generalized ordinary differential equations [1, 5, 6].
The following notations and definitions will be used throughout the pa-per:
R=]− ∞,+∞[;
Rn is a space of real columnn-vectorsx= (xi)n
i=1 with the norm
kxk= n
X
i=1
|xi|;
Rn×n is a space of realn×n-matricesX = (x
ij)ni,j=1 with the norm
kXk= max j=1,...,n
n
X
i=1
|xij|;
IfX∈Rn×n, thenX−1 and det(X) are the matrix inverse toX and the
determinant ofX, respectively;Eis the identityn×nmatrix; b
∨
axand b
∨
aX are the sums of total variations of components of vector- and matrix-functions,x: [a, b]→Rn andX: [a, b]→Rn×n, respectively;
BVn(a, b) is a space of all vector-functions of bounded variation x : [a, b]→Rn (i.e., such that ∨b
ax <+∞) with the norm
kxksup= sup{kx(t)k:t∈[a, b]}1;
x(t−) andx(t+) (x(a−) =x(a),x(b+) =x(b)) are the left and the right limits of the vector-functionx: [a, b]→Rn at the pointt;
d1x(t) =x(t)−x(t−),d2x(t) =x(t+)−x(t);
BVn×n(a, b) is a set of all matrix-functions of bounded variation X : [a, b]→Rn×n, i.e., such that ∨b
aX <+∞;
d1X(t) =X(t)−X(t−),d2X(t) =X(t+)−X(t);
IfX= (xij)ni,j=1∈BVn×n(a, b), thenV(X) : [a, b]→Rn×nis defined by V(X)(a) = 0, V(X)(t) = t
∨
axij
n
i,j=1 (a < t≤b); 1
Ifα∈BV1(a, b),x: [a, b]→Randa≤s < t≤b, then
t
Z
s
x(τ)dα(τ) =x(t)d1α(t) +x(s)d2α(s) +
Z
]s,t[
x(τ)dα(τ),
whereR
]s,t[x(τ)dα(τ) is the Lebesgue–Stieltjes integral over the open
inter-val ]s, t[ (ifs=t, thenRt
sx(τ)dα(τ) = 0);
If A = (aij)ni,j=1 ∈ BVn×n(a, b), X = (xij)ni,j=1 : [a, b] → Rn×n, x =
(xi)n
i=1: [a, b]→Rn anda≤s≤t≤b, then
t
Z
s
dA(τ)·X(τ) = n
X
k=1
t
Z
s
xkj(τ)daik(τ) n
i,j=1,
t
Z
s
dA(τ)·x(τ) = n
X
k=1
t
Z
s
xk(τ)daik(τ)n i=1;
klk is the norm of the linear continuous operatorl:BVn(a, b)→Rn; IfX∈BVn×n(a, b) is the matrix-function with columnsx1, . . . , xn, then
l(X) is the matrix with columnsl(x1), . . . , l(xn).
A functionx∈BVn(a, b) is said to be a solution of problem (1), (2) if it satisfies condition (2) and
x(t) =x(s) + t
Z
s
dA(τ)·x(τ) +f(t)−f(s) for a≤s≤t≤b.
Alongside with (1) and (3), we shall consider the corresponding homoge-neous systems
dx(t) =dA(t)·x(t) (10)
and
dx(t) =dAk(t)·x(t), (30)
respectively.
A matrix-functionY ∈BVn×n(a, b) is said to be a fundamental matrix
of the homogeneous system (10) if
Y(t) =Y(s) + t
Z
s
dA(τ)·Y(τ) for a≤s≤t≤b
and det Y(t)
6
Theorem 1. Let the conditions
det
E+ (−1)j
djA(t)6= 0 for t∈[a, b] (j= 1,2), (6) lim
k→+∞lk(y) =l(y) for y ∈BVn(a, b), (7)
lim
k→+∞ck =c0, (8)
lim
k→+∞supklkk<+∞, (9)
lim k→+∞sup
b
∨
aAk<+∞ (10)
be satisfied and let the conditions
lim k→+∞
Ak(t)−Ak(a)
=A(t)−A(a), (11) lim
k→+∞
fk(t)−fk(a)
=f(t)−f(a) (12)
be fulfilled uniformly on[a, b]. Then for any sufficiently largekproblem(3),
(4)has the unique solution xk and(5)is valid.
To prove the theorem we shall use the following lemmas. Lemma 1. Let αk, βk∈BV1(a, b) (k= 0,1. . .),
lim
k→+∞kβk−β0ksup= 0, (13) r= sup b
∨
a αk :k= 0,1, . . .
<+∞ (14)
and the condition
lim k→+∞
αk(t)−αk(a)
=α0(t)−α0(a) (15) be fulfilled uniformly on[a, b]. Then
lim k→+∞
t
Z
a
βk(τ)dαk(τ) = t
Z
a
β0(τ)dα0(τ)
uniformly on [a, b].
Proof. Letεbe an arbitrary positive number. We denote
Dj(a, b, ε;g) =
t∈[a, b] :djg(t)≥ε
(j= 1,2) where
g(t) =V(β0)(t) for t∈[a, b].
By Lemma 1.1.1 from [5] there exists a finite subdivision
{α0, τ1, α1, . . . , τm, αm} of [a, b] such that
b) Ifτi6∈ D1(a, b, ε;g), theng(τi)−g(αi−1)< ε;
Ifτi∈ D1(a, b, ε;g), thenαi−1< τi andg(τi−)−g(αi−1)< ε;
c) Ifτi6∈ D2(a, b, ε;g), theng(αi)−g(τi)< ε;
Ifτi∈ D2(a, b, ε;g), thenτi< αi andg(αi)−g(τi+)< ε. We set
η(t) =
β0(t) fort∈ {α0, τ1, α1, . . . , τm, αm};
β0(τi−) fort∈]αi−1, τi[, τi∈ D1(a, b, ε;g);
β0(τi) fort∈]αi−1, τi[, τi6∈ D1(a, b, ε;g) or
fort∈]τi, αi[,τi6∈ D2(a, b, ε;g);
β0(τi+) for t∈]τi, αi[,τi∈ D2(a, b, ε;g);
(i= 1, . . . , m).
It can be easily shown thatη∈BV1(a, b) and
|β0(t)−η(t)|<2ε for t∈[a, b]. (16)
For every naturalkandt∈[a, b] we assume
γk(t) = t
Z
a
βk(τ)dαk(t)− t
Z
a
β0(τ)dα0(τ)
and
δk(t) = t
Z
a
η(t)d[αk(τ)−α0(τ)].
It follows from (15) that lim
k→+∞kδkksup= 0. (17)
On the other hand, by (14) and (16) we have
kγkksup≤4rε+rkβk−β0ksup+kδksup (k= 1,2, . . .).
Hence in view of (13) and (17) limk→+∞kγkksup= 0 sinceεis arbitrary.
Lemma 2. Let condition(6) be fulfilled and
lim
k→+∞Yk(t) =Y(t) uniformly on [a, b], (18)
whereY andYk are the fundamental matrices of the homogeneous systems (10)and(30), respectively. Then
inf
det(Y(t))
:t∈[a, b]
>0, (19)
inf
det(Y−1(t))
:t∈[a, b]
and
lim k→+∞Y
−1
k (t) =Y
−1(t) uniformly on [a, b]. (21) Proof. It is known ([6], Theorem III.2.10) thatdjY(t) = djA(t)·Y(t) for
t∈[a, b] (j= 1,2). Therefore (6) implies
det
Y(t−)·Y(t+)
=
det(Y(t))2 ·
2
Y
j=1
det
E+ (−1)j
djA(t)
6
= 0
for t∈[a, b]. (22)
Let us show that (19) is valid. Assume the contrary. Then it can be easily shown that there exists a pointt0∈[a, b] such that detY(t0−)·Y(t0+)=
0. But this equality contradicts (22). Inequality (19) is proved. The proof of inequality (20) is analogous.
In view of (18) and (19) there exists a positive number q such that inf
det(Yk(t))
: t ∈ [a, b]
> q > 0 for any sufficiently large k. From this and (18) we obtain (21).
Proof of the Theorem. Let us show that det
E+ (−1)j
djAk(t)6= 0 for t∈[a, b] (j= 1,2) (23) for any sufficiently largek.
By (11)
lim
k→+∞djAk(t) =djA(t) (j = 1,2) (24)
uniformly on [a, b]. Since ∨b
aA <+∞, the series
P
t∈[a,b]kdjA(t)k(j = 1,2) converges. Thus for anyj∈ {1,2} the inequality
kdjA(t)k ≥ 1 2
may hold only for some finite number of pointstj1, . . . , tj mj in [a, b].
There-fore
kdjA(t)k< 1
2 for t∈[a, b], t6=tji (i= 1, . . . , mj). (25) It follows from (6), (24) and (25) that for any sufficiently large k and for
j∈ {1,2}
det
E+ (−1)j
djAk(tji)6= 0 (i= 1, . . . , mj) (26) and
kdjAk(t)k< 1
The latter inequality implies that the matricesE+ (−1)jd
jAk(t) (j= 1,2) are invertible for t ∈ [a, b], t 6=tji (i = 1, . . . , mj) too. Therefore (23) is proved.
Besides, by (26) and (27) there exists a positive numberr0 such that for
any sufficiently largek
E+ (−1)j
djAk(t)
−1
≤r0 for t∈[a, b] (j = 1,2). (28)
Letkbe a sufficiently large natural number. In view of (6) and (23) there exist ([6], Theorem III.2.10) fundamental matricesY andYk of systems (10)
and (30), respectively, satisfying Y(a) = Yk(a) = E. Moreover, Yk−1 ∈
BVn×n(a, b).
Let us prove (18). We setZk(t) =Yk(t)−Y(t) fort∈[a, b] andBk(t) =
Ak(t−) fort∈[a, b]. Obviously, for everyt∈[a, b]
d1Bk(t)−Ak(t)=−d2Bk(t)−Ak(t)=−d1Ak(t) and
t
Z
a
d
Bk(τ)−Ak(τ)·Zk(τ) =−d1Ak(t)·Zk(t).
Consequently,
Zk(t)≡
E−d1Ak(t)
−1h
t
Z
a
d
Ak(τ)−A(τ)
·Y(τ) + t
Z
a
dBk(τ)·Zk(τ)
i .
From this and (28) we get
kZk(t)k ≤r0
εk+
t
Z
a
dkV(Bk)(τ)k · kZk(τ)k
for t∈[a, b],
where
εk= sup
n
t
Z
a
d
Ak(τ)−A(τ)
·Y(τ)
:t∈[a, b] o
.
Hence, according to the Gronwall inequality ([6], Theorem I.4.30),
kZk(t)k ≤r0εkexp
r0
b
∨
a Bk
≤r0εkexp
r0
b
∨
a Ak
for t∈[a, b].
By (10), (11) and Lemma 1 this inequality implies (18).
It is known ([6], Theorem III.2.13) that ifxk is the solution of (3), then
xk(t)≡Yk(t)xk(a) +fk(t)−fk(a)−Yk(t) t
Z
a
dYk−1(τ)·
fk(τ)−fk(a)
Thus problem (3), (4) has the unique solution if and only if det
lk(Yk)
6
= 0. (29)
Since problem (1), (2) has the unique solutionx0, we have
det l(Y)
6
= 0. (30)
Besides, by (7), (9) and (18) lim
k→+∞lk(Yk) =l(Y).
Therefore, in view of (30), there exists a natural numberk0such that
condi-tion (29) is fulfilled for everyk≥k0. Thus problem (3), (4) has the unique
solutionxk fork≥k0 and
xk(t)≡Yk(t) lk(Yk)
−1
ck−lk(Fk(fk))+Fk(fk)(t), (31) where
Fk(fk)(t) =fk(t)−fk(a)−Yk(t) t
Z
a
dYk−1(τ)·
fk(τ)−fk(a).
According to Lemma 2 condition (21) is fulfilled and
ρ= sup kY−1
k (t)k+kYk(t)k:t∈[a, b], k≥k0
<+∞. (32) The equality
Yk−1(t)−Yk−1(s) =Yk−1(s) s
Z
t
dAk(τ)·Yk(τ) Yk−1(t)
implies
kYk−1(t)−Y
−1
k (s)k ≤ρ
3∨t
sAk for a≤s≤t≤b (k≥k0). This inequality, together with (10) and (32), yields
lim k→+∞sup
b
∨
aY
−1
k <+∞. By this, (12) and (21) it follows from Lemma 1 that
lim k→+∞
t
Z
a
dYk−1(τ)·
fk(τ)−fk(a)
=
= t
Z
a
dY−1(τ)·
f(τ)−f(a)
uniformly on [a, b].
Using (7)-(9), (12), (18), (29), (30) and (33), from (31) we get lim k→+∞xk(t)
=z(t) uniformly on [a, b], where
z(t) =Y(t)
l(Y)−1
c0−lF(f)+F(f)(t),
F(f)(t) =f(t)−f(a)−Y(t) t
Z
a
dY−1(τ)·
f(τ)−f(a) .
It is easy to verify that the vector-functionz: [a, b]→Rn is the solution of problem (1), (2). Therefore
x0(t) =z(t) for t∈[a, b].
References
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(Received 2.07.1993) Author’s address:
I. Vekua Institute of Applied Mathematics of Tbilisi State University